Question

attempt 1: 10 attempts remaining. find $f(4)$ if $f(x)=\frac{3x^{2}+5x + 2}{sqrt{x}}$. $f(4)=$ (round to two decimal places as needed.) submit answer next item

Explanation:

Step1: Rewrite the function

Rewrite $f(x)=\frac{3x^{2}+5x + 2}{\sqrt{x}}$ as $f(x)=3x^{\frac{3}{2}}+5x^{\frac{1}{2}}+2x^{-\frac{1}{2}}$.

Step2: Differentiate term - by - term

Using the power rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, we have $f'(x)=3\times\frac{3}{2}x^{\frac{3}{2}-1}+5\times\frac{1}{2}x^{\frac{1}{2}-1}+2\times(-\frac{1}{2})x^{-\frac{1}{2}-1}=\frac{9}{2}x^{\frac{1}{2}}+\frac{5}{2}x^{-\frac{1}{2}}-x^{-\frac{3}{2}}$.

Step3: Evaluate $f'(4)$

Substitute $x = 4$ into $f'(x)$:
\[

$$\begin{align*} f'(4)&=\frac{9}{2}\sqrt{4}+\frac{5}{2}\frac{1}{\sqrt{4}}-\frac{1}{(\sqrt{4})^{3}}\\ &=\frac{9}{2}\times2+\frac{5}{2}\times\frac{1}{2}-\frac{1}{8}\\ &=9+\frac{5}{4}-\frac{1}{8}\\ &=9+\frac{10 - 1}{8}\\ &=9+\frac{9}{8}\\ &=9 + 1.125\\ &=10.125 \end{align*}$$

\]
Rounding to two decimal places, we get $10.13$.

Answer:

$10.13$